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Chapter 1: Problem 21

Recall the definition of the real arccos: arccos is the inverse function of \(\cos\) restricted as a function \([0, \pi] \rightarrow[-1,1]\), thus $$ \arccos t=\varphi \quad \Longleftrightarrow \quad 0 \leq \varphi \leq \pi \text { and } \cos \varphi=t $$ Show: For \(z=x+\mathrm{i} y \neq 0\) we have $$ \operatorname{Arg} z= \begin{cases}\pi & \text { if } y=0 \text { and } x<0 \\\ \operatorname{sgn}(y) \arccos \frac{x}{\sqrt{x^{2}+y^{2}}} & \text { otherwise. }\end{cases} $$

Short Answer

Expert verified
The expression for \( \operatorname{Arg} z \) is valid based on cases given for \( y = 0, x < 0 \) and otherwise.

Step by step solution

01

Understand the Problem

The task is to show that the argument \( \operatorname{Arg} z \) of a complex number \( z = x + \mathrm{i} y \) is given by specific formulas, depending on the conditions provided. These conditions are related to the sign of \( y \) and the value of \( x \).
02

Recall the Definition of Arg z

The argument \( \operatorname{Arg} z \) of a complex number \( z = x + \mathrm{i} y \) is the angle \( \theta \) such that \( z = r (\cos \theta + \mathrm{i} \sin \theta) \), where \( r = |z| = \sqrt{x^2 + y^2} \). The value of the argument is typically between \( -\pi \) and \( \pi \).
03

Case 1 - If y=0 and x

For \( y = 0 \) and \( x < 0 \), the point \( z = x \) lies on the negative real axis. Therefore, the angle corresponding to \( z \) is \( \pi \), since the angle in the positive direction from 0 on a unit circle to this point is \( \pi \).
04

Case 2 - Otherwise

For any other case, we use the formula \( \operatorname{Arg} z = \operatorname{sgn}(y) \arccos \frac{x}{\sqrt{x^{2}+y^{2}}} \). Here, \( \arccos \frac{x}{\sqrt{x^2 + y^2}} \) gives the angle in the range \([0, \pi]\), and \( \operatorname{sgn}(y) \) adjusts the angle to the correct sign. \( \operatorname{sgn}(y) \) is \(+1\) if \( y > 0\), \(-1\) if \( y < 0\), ensuring the angle is properly directed.
05

Confirm by Example

To further confirm, consider an example where \( z=1+\mathrm{i} \) (i.e., \( x=1, y=1 \)). The modulus \( \sqrt{1^2 + 1^2} = \sqrt{2} \) leads to \( \cos \theta = \frac{1}{\sqrt{2}} \), which means \( \theta = \arccos \left( \frac{1}{\sqrt{2}} \right) = \frac{\pi}{4} \). The sign of \( y \) is positive, so the argument \( \operatorname{Arg} z = \frac{\pi}{4} \), which matches our computation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Inverse Trigonometric Functions
Inverse trigonometric functions are calculations that take values and return angles. These functions essentially reverse the process of standard trigonometric functions. Consider the arccosine or \( \arccos \). It takes a value within \([-1, 1]\) and finds the angle in radians between \([0, \pi]\) whose cosine is that value.
Understanding \( \arccos \) is crucial because it helps us work with angles when dealing with projections onto the real axis. For example, in the step-by-step solution, the formula \( \operatorname{Arg} z = \operatorname{sgn}(y) \arccos \frac{x}{\sqrt{x^2+y^2}} \) involves \( \arccos \). It determines the angle between a vector and the positive real axis, but adjusts direction using the sign of \( y \).
  • \( \arccos \left( \frac{x}{\sqrt{x^2 + y^2}} \right) \) gives the uncoded angle based on length comparisons.
  • It returns angles that help us bind our calculations to real, calculable quantities.
  • The range \([0, \pi]\) ensures every angle is unique for a given projection.
Complex Numbers
Complex numbers expand our basic understanding of numbers by incorporating the imaginary unit, denoted as \( \mathrm{i} \), where \( \mathrm{i}^2 = -1 \). These numbers are written as \( z = x + \mathrm{i} y \), with \( x \) as the real part and \( y \) as the imaginary part.
They allow us to perform computations and represent phenomena in a more comprehensive way, particularly useful in advanced mathematics and engineering.
  • Complex numbers are plotted on a plane, called the complex plane, which has a horizontal axis (Real) and a vertical axis (Imaginary).
  • The modulus \( |z| = \sqrt{x^2 + y^2} \) describes the distance from the origin to the point \( z \).
  • They provide a complete extension to solve problems regular real numbers cannot, like square roots of negative numbers.
Understanding these properties helps in grasping the significance and application of the argument of a complex number 鈥� which measures a number鈥檚 direction.
Argument of a Complex Number
The argument, \( \operatorname{Arg} z \), of a complex number is the angle formed with the positive real axis on the complex plane. It gives the direction or angle \( \theta \) and helps in transitioning complex numbers to polar form.
Essentially, the argument tells you the direction of a line connecting the origin to the point \( z \). This direction is often considered between \(-\pi\) and \(\pi\).
A critical example involves determining if \( y = 0 \) and \( x < 0 \). Here, the argument is precisely \( \pi \) because the point lies on the negative real-axis, hence making the argument along the circular plane half of the circle.
  • The formula \( \operatorname{Arg} z = \operatorname{sgn}(y) \arccos \frac{x}{\sqrt{x^2 + y^2}} \) covers cases for both positive and negative y-coordinates, allowing us to calculate direction accurately.
  • Understanding the argument helps unlock the geometrical perspective needed for visualizing complex numbers on the plane.
  • This concept is also key when transforming complex numbers into their exponential form \( z = re^{\mathrm{i}\theta} \).
Trigonometry
Trigonometry deals with relationships between the angles and sides of triangles, typically in a circle. This section defines the essential links needed for complex analysis when working with complex numbers and their geometric representations.
In the exercise, resolving \( \cos \) and \( \arccos \) functions required understanding trigonometry thoroughly. The standard cosine function \( \cos \theta = \frac{x}{\sqrt{x^2+y^2}} \) originates from the relationship of sides and hypotenuse in a right triangle.
Trigonometry simplifies finding angles and provides a way to use them for rotational points on a plane.
  • It bridges the gap between algebraic expressions and geometric interpretations of complex numbers.
  • Angles derived using trigonometry are integral in calculations across various domains, such as physics and engineering.
  • Therefore, mastering basic trigonometric functions is fundamental to effectively grasping complex analysis concepts.
Improved comprehension of these angles and projections helps define or redefine trigonometric functions' various parameters and their inverses effectively.

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Most popular questions from this chapter

Assume the series \(\sum a_{n}\) is absolutely convergent, and let \(a:=\sum_{n=0}^{\infty} a_{n} .\) Suppose the series \(\sum b_{n}\) is convergent, and assume \(b:=\sum_{n=0}^{\infty} b_{n}\) Show: if \(c_{n}:=\sum_{\nu=0}^{n} a_{\nu} b_{n-\nu}\), then the series \(\sum c_{n}\) is convergent, and for \(c:=\) \(\sum_{n=0}^{\infty} c_{n}\) one has \(c=a b \quad\) (MERTENS' theorem, F. MERTENS, 1875\()\)

Let \(a \in \mathbb{C}^{\bullet}\) be given. For which \(z_{0} \in \mathbb{C}\) converges the series \(\left(z_{n}\right)\), which is recursively defined by $$ z_{n+1}=\frac{1}{2}\left(z_{n}+\frac{a}{z_{n}}\right) \quad \text { for } n \geq 0 \quad ? $$ Remark: Both exercises 1 and 2 are special complex instances of NEWTON's approximation method for zeros (of the polynomials \(\left.z^{2}-a\right) .\) See also Exercise 7 in I.4.

Let \(\operatorname{Maps}\left(\mathbb{N}_{0}, \mathbb{C}\right)\) be the set of all maps of \(\mathbb{N}_{0}\) into \(\mathbb{C}(=\) the set of all complex number sequences). Show: The map $$ \begin{aligned} \sum: \operatorname{Maps}\left(\mathbb{N}_{0}, \mathbb{C}\right) & \longrightarrow \operatorname{Maps}\left(\mathbb{N}_{0}, \mathbb{C}\right) \\ \left(a_{n}\right)_{n} \geq 0 & \longmapsto\left(S_{n}\right)_{n \geq 0} \text { with } S_{n}:=a_{0}+a_{1}+\cdots+a_{n} \end{aligned} $$ is bijective the (telescope trick). The theories of sequences and of infinite series are therefore in principle the same.

Determine, in each case, all the \(z \in \mathbb{C}\) with $$ \begin{aligned} \exp (z) &=-2, & \exp (z) &=\mathrm{i}, & & \exp (z) &=-\mathrm{i} \\ \sin z &=100, & \sin z &=7 \mathrm{i}, & \sin z &=1-\mathrm{i} \\ \cos z &=3 \mathrm{i}, & \cos z &=3+4 \mathrm{i}, & \cos z &=13. \end{aligned} $$

Determine all the harmonic functions $$ u: \mathbb{C}^{\bullet}=\mathbb{R}^{2} \backslash\\{(0,0)\\} \longrightarrow \mathbb{R} $$ that depend only on \(r:=\sqrt{x^{2}+y^{2}} .\)
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